在列上使用 starts_with 和 if_else 的 mutate
Use mutate with starts_with on columns and if_else
我有多个 Tibbles，我想成为
用一个函数改变它们。
问题是列名部分不同：
这就是我要申请的
mpg %>% mutate(model_QQ = if_else(year == 1999, 0 , if_else(cty > 20, 1, -1)))
但是 tibbles 的同名如下：
c("audi","nissan","subaru") %>% set_names() %>% map(~mpg %>% rename_at(vars(model:class), list(~str_c(.,"_",!!quo(.x)))))
我正在考虑做类似的东西：
mpg %>% mutate(model_QQ = if_else(starts_with("year") == 1999, 0 , if_else(starts_with("cty") > 20, 1, -1)))
但这些仅在 select 函数内部起作用。
有什么建议吗？
EDIT
I remake the code with case when, maybe is made more clear that way about what I'm looking for:
mpg %>%
mutate(
model_QQ = case_when(
starts_with("year") == 1999 ~ 0L,
starts_with("cty") > 20 ~ 1L,
starts_with("cty") <= 20 ~ -1L
))
你可以这样做：
mpg %>%
mutate(across(starts_with("year"), ~if_else(.x >= 1999, 0 , 1)),
across(starts_with("cty"), ~if_else(.x > 20, 1, -1)))
#> # A tibble: 234 x 11
#> manufacturer model displ year cyl trans drv cty hwy fl class
#> <chr> <chr> <dbl> <dbl> <int> <chr> <chr> <dbl> <int> <chr> <chr>
#> 1 audi a4 1.8 0 4 auto(~ f -1 29 p comp~
#> 2 audi a4 1.8 0 4 manua~ f 1 29 p comp~
#> 3 audi a4 2 0 4 manua~ f -1 31 p comp~
#> 4 audi a4 2 0 4 auto(~ f 1 30 p comp~
#> 5 audi a4 2.8 0 6 auto(~ f -1 26 p comp~
#> 6 audi a4 2.8 0 6 manua~ f -1 26 p comp~
#> 7 audi a4 3.1 0 6 auto(~ f -1 27 p comp~
#> 8 audi a4 qu~ 1.8 0 4 manua~ 4 -1 26 p comp~
#> 9 audi a4 qu~ 1.8 0 4 auto(~ 4 -1 25 p comp~
#> 10 audi a4 qu~ 2 0 4 manua~ 4 -1 28 p comp~
#> # ... with 224 more rows
或者，如果您希望将它们重命名为 year_modelQQ 和 cty_modelQQ，您可以这样做：
mpg %>%
mutate(across(starts_with("year"), list(modelQQ = ~if_else(.x >= 1999, 0 ,1))),
across(starts_with("cty"), list(modelQQ = ~if_else(.x > 20, 1, -1))))
#> # A tibble: 234 x 13
#> manufacturer model displ year cyl trans drv cty hwy fl class year_modelQQ
#> <chr> <chr> <dbl> <int> <int> <chr> <chr> <int> <int> <chr> <chr> <dbl>
#> 1 audi a4 1.8 1999 4 auto~ f 18 29 p comp~ 0
#> 2 audi a4 1.8 1999 4 manu~ f 21 29 p comp~ 0
#> 3 audi a4 2 2008 4 manu~ f 20 31 p comp~ 0
#> 4 audi a4 2 2008 4 auto~ f 21 30 p comp~ 0
#> 5 audi a4 2.8 1999 6 auto~ f 16 26 p comp~ 0
#> 6 audi a4 2.8 1999 6 manu~ f 18 26 p comp~ 0
#> 7 audi a4 3.1 2008 6 auto~ f 18 27 p comp~ 0
#> 8 audi a4 q~ 1.8 1999 4 manu~ 4 18 26 p comp~ 0
#> 9 audi a4 q~ 1.8 1999 4 auto~ 4 16 25 p comp~ 0
#> 10 audi a4 q~ 2 2008 4 manu~ 4 20 28 p comp~ 0
#> # ... with 224 more rows, and 1 more variable: cty_modelQQ <dbl>
编辑
根据更新的信息，这应该可以解决问题：
mpg %>%
mutate(across(starts_with("year"), list(A = ~1 - (.x == 1999))),
across(starts_with("cty"), list(A = ~ -1 + 2*(.x > 20))),
modelQQ = ifelse(year_A == 0, 0, cty_A)) %>%
select(-ends_with("_A")
我们可以使用 case_when
library(dplyr)
mpg %>%
mutate(across(starts_with("year"), ~case_when(.x == 1999 ~ 0, TRUE ~ 1)),
across(starts_with("cty"), ~case_when(.x > 20 ~ 1, TRUE ~ -1)))
如果是创建单列
mpg %>%
mutate(model_QQ = case_when(select(., starts_with("year")) ==
1999 ~ 0L,
select(., starts_with('cty')) > 20 ~ 1L, TRUE ~ -1L))
-输出
# A tibble: 234 x 12
# manufacturer model displ year cyl trans drv cty hwy fl class model_QQ
# <chr> <chr> <dbl> <int> <int> <chr> <chr> <int> <int> <chr> <chr> <int>
# 1 audi a4 1.8 1999 4 auto(l5) f 18 29 p compact 0
# 2 audi a4 1.8 1999 4 manual(m5) f 21 29 p compact 0
# 3 audi a4 2 2008 4 manual(m6) f 20 31 p compact -1
# 4 audi a4 2 2008 4 auto(av) f 21 30 p compact 1
# 5 audi a4 2.8 1999 6 auto(l5) f 16 26 p compact 0
# 6 audi a4 2.8 1999 6 manual(m5) f 18 26 p compact 0
# 7 audi a4 3.1 2008 6 auto(av) f 18 27 p compact -1
# 8 audi a4 quattro 1.8 1999 4 manual(m5) 4 18 26 p compact 0
# 9 audi a4 quattro 1.8 1999 4 auto(l5) 4 16 25 p compact 0
#10 audi a4 quattro 2 2008 4 manual(m6) 4 20 28 p compact -1
# … with 224 more rows
我有多个 Tibbles，我想成为 用一个函数改变它们。 问题是列名部分不同： 这就是我要申请的
mpg %>% mutate(model_QQ = if_else(year == 1999, 0 , if_else(cty > 20, 1, -1)))
但是 tibbles 的同名如下：
c("audi","nissan","subaru") %>% set_names() %>% map(~mpg %>% rename_at(vars(model:class), list(~str_c(.,"_",!!quo(.x)))))
我正在考虑做类似的东西：
mpg %>% mutate(model_QQ = if_else(starts_with("year") == 1999, 0 , if_else(starts_with("cty") > 20, 1, -1)))
但这些仅在 select 函数内部起作用。 有什么建议吗？
EDIT I remake the code with case when, maybe is made more clear that way about what I'm looking for:
mpg %>%
mutate(
model_QQ = case_when(
starts_with("year") == 1999 ~ 0L,
starts_with("cty") > 20 ~ 1L,
starts_with("cty") <= 20 ~ -1L
))
你可以这样做：
mpg %>%
mutate(across(starts_with("year"), ~if_else(.x >= 1999, 0 , 1)),
across(starts_with("cty"), ~if_else(.x > 20, 1, -1)))
#> # A tibble: 234 x 11
#> manufacturer model displ year cyl trans drv cty hwy fl class
#> <chr> <chr> <dbl> <dbl> <int> <chr> <chr> <dbl> <int> <chr> <chr>
#> 1 audi a4 1.8 0 4 auto(~ f -1 29 p comp~
#> 2 audi a4 1.8 0 4 manua~ f 1 29 p comp~
#> 3 audi a4 2 0 4 manua~ f -1 31 p comp~
#> 4 audi a4 2 0 4 auto(~ f 1 30 p comp~
#> 5 audi a4 2.8 0 6 auto(~ f -1 26 p comp~
#> 6 audi a4 2.8 0 6 manua~ f -1 26 p comp~
#> 7 audi a4 3.1 0 6 auto(~ f -1 27 p comp~
#> 8 audi a4 qu~ 1.8 0 4 manua~ 4 -1 26 p comp~
#> 9 audi a4 qu~ 1.8 0 4 auto(~ 4 -1 25 p comp~
#> 10 audi a4 qu~ 2 0 4 manua~ 4 -1 28 p comp~
#> # ... with 224 more rows
或者，如果您希望将它们重命名为 year_modelQQ 和 cty_modelQQ，您可以这样做：
mpg %>%
mutate(across(starts_with("year"), list(modelQQ = ~if_else(.x >= 1999, 0 ,1))),
across(starts_with("cty"), list(modelQQ = ~if_else(.x > 20, 1, -1))))
#> # A tibble: 234 x 13
#> manufacturer model displ year cyl trans drv cty hwy fl class year_modelQQ
#> <chr> <chr> <dbl> <int> <int> <chr> <chr> <int> <int> <chr> <chr> <dbl>
#> 1 audi a4 1.8 1999 4 auto~ f 18 29 p comp~ 0
#> 2 audi a4 1.8 1999 4 manu~ f 21 29 p comp~ 0
#> 3 audi a4 2 2008 4 manu~ f 20 31 p comp~ 0
#> 4 audi a4 2 2008 4 auto~ f 21 30 p comp~ 0
#> 5 audi a4 2.8 1999 6 auto~ f 16 26 p comp~ 0
#> 6 audi a4 2.8 1999 6 manu~ f 18 26 p comp~ 0
#> 7 audi a4 3.1 2008 6 auto~ f 18 27 p comp~ 0
#> 8 audi a4 q~ 1.8 1999 4 manu~ 4 18 26 p comp~ 0
#> 9 audi a4 q~ 1.8 1999 4 auto~ 4 16 25 p comp~ 0
#> 10 audi a4 q~ 2 2008 4 manu~ 4 20 28 p comp~ 0
#> # ... with 224 more rows, and 1 more variable: cty_modelQQ <dbl>
编辑
根据更新的信息，这应该可以解决问题：
mpg %>%
mutate(across(starts_with("year"), list(A = ~1 - (.x == 1999))),
across(starts_with("cty"), list(A = ~ -1 + 2*(.x > 20))),
modelQQ = ifelse(year_A == 0, 0, cty_A)) %>%
select(-ends_with("_A")
我们可以使用 case_when
library(dplyr)
mpg %>%
mutate(across(starts_with("year"), ~case_when(.x == 1999 ~ 0, TRUE ~ 1)),
across(starts_with("cty"), ~case_when(.x > 20 ~ 1, TRUE ~ -1)))
如果是创建单列
mpg %>%
mutate(model_QQ = case_when(select(., starts_with("year")) ==
1999 ~ 0L,
select(., starts_with('cty')) > 20 ~ 1L, TRUE ~ -1L))
-输出
# A tibble: 234 x 12
# manufacturer model displ year cyl trans drv cty hwy fl class model_QQ
# <chr> <chr> <dbl> <int> <int> <chr> <chr> <int> <int> <chr> <chr> <int>
# 1 audi a4 1.8 1999 4 auto(l5) f 18 29 p compact 0
# 2 audi a4 1.8 1999 4 manual(m5) f 21 29 p compact 0
# 3 audi a4 2 2008 4 manual(m6) f 20 31 p compact -1
# 4 audi a4 2 2008 4 auto(av) f 21 30 p compact 1
# 5 audi a4 2.8 1999 6 auto(l5) f 16 26 p compact 0
# 6 audi a4 2.8 1999 6 manual(m5) f 18 26 p compact 0
# 7 audi a4 3.1 2008 6 auto(av) f 18 27 p compact -1
# 8 audi a4 quattro 1.8 1999 4 manual(m5) 4 18 26 p compact 0
# 9 audi a4 quattro 1.8 1999 4 auto(l5) 4 16 25 p compact 0
#10 audi a4 quattro 2 2008 4 manual(m6) 4 20 28 p compact -1
# … with 224 more rows